Power Series of Integral xarctan(3x)

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The integral of xarctan(3x) from 0 to 0.1 is evaluated by expressing it as a power series. The user differentiated the function to obtain power series for arctan(3x) and x/(1+9x^2), then integrated these series. The resulting evaluation yielded 0.0050927, which was identified as incorrect. A suggestion was made to verify the differentiation process and to note that the integral of x/(1+9x^2) results in a logarithmic function, not arctan. The discussion highlights the importance of careful differentiation and integration in power series evaluation.
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Homework Statement



Evaluate the integral of xarctan(3x) from 0 to 0.1 by expressing the integral in terms of a power series.

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The Attempt at a Solution



I differentiated xarctan(3x) until I got two functions that I could turn into power series (arctan(3x) and x(1/(1+9x^2)).

After I found the power series of these two functions I integrated them as many times as was necessary to arrive back at the integral of the original function and got (both as sums from one to infinity):
[(-1)^(n-1)*9^(n-1)*x^(2n+1)]/[(2n+1)(2n)(2n-1)] + [(-1)^(n-1)*9^(n-1)*x^(2n)]/[(2n)(2n-1)]

When I evaluated it from 0 to 0.1, I ended up with 0.0050927 which is not the correct answer, but I don't see where I would have done it wrong. Any help would be appreciated.
 
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A faster way to find the power series would be knowing \arctan{u}=\sum_{n=0}^{\infty}(-1)^{n}\frac{u^{2n+1}}{2n+1} convergent for |x|\leq{1}. x\arctan{3x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{2n+1}}{2n+1}x^{2(n+1)}. I haven't checked your work but maybe you made a mistake while differentiating.

On a separate note, \int \frac{x}{1+9x^{2}}\,dx=\frac{1}{18}\ln{|1+9x^{2}|}+C, not the arctan you were looking for.
 
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